What is the reason TypeScript struggles to automatically deduce assignments of identical object types?

Question

What is the reason TypeScript struggles to automatically deduce assignments of identical object types?

Imagine a scenario with a simple code snippet to illustrate the issue:

interface I {
  n?: number;
  s?: string;
}

const a: I = {
  n: 1,
}

const b: I = {
  n: 2,
  s: 'b',
}

const props = ['n', 's'] as const;

for (const prop of props) {
  if (!a[prop]) {
    // encountering an error:
    // TS2322: Type 'string | number' is not assignable to type 'never'.
    //   Type 'string' is not assignable to type 'never'.
    a[prop] = b[prop];
  }
}

If a and b are of the same type, accessing the same property prop, shouldn't it work smoothly?

If b[prop] holds a number, then a[prop] should accept numbers
If b[prop] contains a string, then a[prop] should accept strings
If b[prop] is undefined, it implies that the prop is optional?

So, what could be the missing piece here?

Update: After simplifying the example too much, the fix was to eliminate the as const part... However, this behavior might be specific to my setup since I'm utilizing strict: true in tsconfig.json...

When attempting access without as const, I encounter the TS7053 error:

Element implicitly has an 'any' type because expression of type 'string' can't be used to index type 'I'.

No index signature with a parameter of type 'string' was found on type 'I'.

To resolve this, add the as const or use

for(const prop of props as ('n' | 's')[]) {

Later on, when faced with the original error (easily resolved by adding as any but aiming to avoid it)

typescript

Answer 1

Answer №1

Typescript does not verify every possible property of your code, as it is not its intention to do so. In this specific scenario, the relationship between b[prop] and a[prop] in terms of type matching may seem obvious to us humans because the property names are identical. However, Typescript lacks a rule that would allow it to recognize this similarity.

Following Typescript's logic:

props is defined as a tuple with the types 'n' and 's'.
The type of prop is inferred as 'n' | 's' since it was declared as an element of props.

Given that b is of type I, b[prop] is then inferred as

I['n' | 's']</code which simplifies to <code>string | number

.

Considering that a is also of type I, the assignment target a[prop] can only accept a value that is assignable to any property that prop might refer to; hence the inferred type becomes
```
I['n'] & I['s']</code leading to <code>string & number
```
, and ultimately never.
From Typescript's perspective, the assignment a[prop] = b[prop] attempts to assign a value of type string | number to a target of type never, resulting in a type error.

A simple solution is to utilize a type assertion such as b[prop] as any to inform Typescript that the code is indeed type-safe and does not require verification. Alternatively, you could employ a helper function that handles assignments in a manner compliant with Typescript's type-checker:

function copyMissingProperties<T>(a: T, b: T, props: readonly (keyof T)[]): void {
    for (const prop of props) {
        if (!a[prop]) {
            a[prop] = b[prop]; // acceptable
        }
    }
}

In this context, the type of prop is keyof T, eliminating the need for an intersection type in the assignment target.

Answer 2

Typescript does not verify every possible property of your code, as it is not its intention to do so. In this specific scenario, the relationship between b[prop] and a[prop] in terms of type matching may seem obvious to us humans because the property names are identical. However, Typescript lacks a rule that would allow it to recognize this similarity.

Following Typescript's logic:

props is defined as a tuple with the types 'n' and 's'.
The type of prop is inferred as 'n' | 's' since it was declared as an element of props.

Given that b is of type I, b[prop] is then inferred as

I['n' | 's']</code which simplifies to <code>string | number

.

Considering that a is also of type I, the assignment target a[prop] can only accept a value that is assignable to any property that prop might refer to; hence the inferred type becomes
```
I['n'] & I['s']</code leading to <code>string & number
```
, and ultimately never.
From Typescript's perspective, the assignment a[prop] = b[prop] attempts to assign a value of type string | number to a target of type never, resulting in a type error.

A simple solution is to utilize a type assertion such as b[prop] as any to inform Typescript that the code is indeed type-safe and does not require verification. Alternatively, you could employ a helper function that handles assignments in a manner compliant with Typescript's type-checker:

function copyMissingProperties<T>(a: T, b: T, props: readonly (keyof T)[]): void {
    for (const prop of props) {
        if (!a[prop]) {
            a[prop] = b[prop]; // acceptable
        }
    }
}

In this context, the type of prop is keyof T, eliminating the need for an intersection type in the assignment target.

Answer 3

Answer №2

Here's a clever workaround I found that doesn't rely on casting:

interface I {
    n?: number;
    s?: string;
}

const a: I = {
    n: 1,
}

const b: I = {
    n: 2,
    s: 'b',
}

const props: Array<keyof I> = ['n', 's'];

for (const prop of props) {
    if (!a[prop]) {
        const newProp: I = {[prop]: b[prop]};
        Object.assign(a, newProp);
    }
}

The key was using Object.assign. To maintain type safety, define the new property as an object of type I and then utilize it in Object.assign. For a universal solution, simply use

Object.assign(a, {[prop]: b[prop]})

Answer 4

Here's a clever workaround I found that doesn't rely on casting:

interface I {
    n?: number;
    s?: string;
}

const a: I = {
    n: 1,
}

const b: I = {
    n: 2,
    s: 'b',
}

const props: Array<keyof I> = ['n', 's'];

for (const prop of props) {
    if (!a[prop]) {
        const newProp: I = {[prop]: b[prop]};
        Object.assign(a, newProp);
    }
}

The key was using Object.assign. To maintain type safety, define the new property as an object of type I and then utilize it in Object.assign. For a universal solution, simply use

Object.assign(a, {[prop]: b[prop]})

Answer 5

Answer №3

The issue lies within this particular line of code

const props = ['n', 's'] as const;

Please update it to

const props = ['n', 's'];

For further reference, you can view this StackBlitz project

Answer 6

The issue lies within this particular line of code

const props = ['n', 's'] as const;

Please update it to

const props = ['n', 's'];

For further reference, you can view this StackBlitz project

Answer 7

Answer №4

I successfully tested this approach and it worked well. It is important to note that the const value should be removed from props definition.

interface I {
  n?: number;
  s?: string;
}

const a: I = {
  n: 1,
}

const b: I = {
  n: 2,
  s: 'b',
}

const props = ['n', 's'];

for (const prop of props) {
  if (!a[prop]) {
    a[prop] = b[prop];
    console.log(a, b);
  }
}

The final output will be: {n: 1, s: "b"} {n: 2, s: "b"}

Answer 8

I successfully tested this approach and it worked well. It is important to note that the const value should be removed from props definition.

interface I {
  n?: number;
  s?: string;
}

const a: I = {
  n: 1,
}

const b: I = {
  n: 2,
  s: 'b',
}

const props = ['n', 's'];

for (const prop of props) {
  if (!a[prop]) {
    a[prop] = b[prop];
    console.log(a, b);
  }
}

The final output will be: {n: 1, s: "b"} {n: 2, s: "b"}

What is the reason TypeScript struggles to automatically deduce assignments of identical object types?

Answer №1

Answer №2

Answer №3

Answer №4

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